Proposition 2.1.8

  1. If is an invertible element in , then so is , and belongs to . Clearly, .
    The normalization of an invertible element is unitary.
  2. The map defined in 1. is continuous for every , and in for every .
    Normalizing is continuous over the unitaries, moreover, the normalization of z is homotopic to z in the set of invertible of A.
  3. If are unitary elements in , and in , then in
    If two unitaries are homotopic in the invertibles, we can strengthen this condition to homotopic inside the unitaries.
    Proof of 1:
    If then and by closure
    It follows that is invertible. (WHY?_Perhaps the continuity of the square root?)
    Put
    Then Moreover is invertible, since it is a product of two invertible elements and , since
    Proof of 2:

Multiplication in a -algebra is continuous, and so is the map in . To show that is continuous, it therefore suffices to show that the map is continuous. This map is composed of the map and the map for .
The former of these maps is continuous because the involution and multiplication is continuous. It now suffices to show that is continuous on any bounded set. But this follows from Lemma 1.2.5. because each bounded subset of is contained in , where and
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Aside:
is the set of self-adjoints in with spectrum contained in .
Lemma 1.2.5. Says that any continuous function on , then is continuous.
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If is a unitary element in , then and so .
Let be given, and put for .This is the homotopy between z and its normalization
Then and . Since is positive and invertible, there exists for which .
For each , we have
Thus and are invertible. The map is continuous, and so in

Proof of 3: If is a continuous path in from to , then is a continuous path in from to .